Answer : The 
for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : 
Reduction : 
Now we have to calculate the Gibbs free energy.
Formula used :
where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the for this reaction is, -88780 J/mole.
<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15%
Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this
fertilizer, to get the number of moles of phosphorus, you multiply the mass by
35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the
calculated mass of phosphorous by its molar mass which is 30.97 g/mol.
Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>
Answer:
20cm^2
Explanation:
Here, Density= Mass/ Volume
=100/5
= 20 cm^2
Answer:
8.10 hours.
Explanation:
You start with 500.0g.
After the first half-life, you have 250.0g.
After the second, you have 125.0g.
After the third, you have 62.50g.
Therefore, it takes three half-lives to decay to 62.50g.
Therefore, the elapsed time must be triple the length of one half-life.
24.3
3
=
8.10
, so it is 8.10 hours.
Explanation:
You would need 1000 liters
